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leetcode - Intersection of Two Linked Lists
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.
Notes:
null.1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { 12 if(headA == NULL || headB == NULL){ 13 return NULL; 14 } 15 ListNode *tpA = headA; 16 ListNode *tpB = headB; 17 int numA=0,numB=0; 18 while(tpA != NULL){ 19 tpA = tpA->next; 20 numA++; 21 } 22 while(tpB != NULL){ 23 tpB = tpB->next; 24 numB++; 25 } 26 tpA = headA; 27 tpB = headB; 28 if(numA!=numB){ 29 if(numA>numB){ 30 for( int i = 0; i<(numA-numB); i++){ 31 tpA = tpA->next; 32 } 33 } 34 else{ 35 for( int i = 0; i<(numB-numA); i++){ 36 tpB = tpB->next; 37 } 38 } 39 } 40 while(tpA != NULL){ 41 if(tpA == tpB){ 42 return tpA; 43 } 44 else{ 45 tpA = tpA->next; 46 tpB = tpB->next; 47 } 48 } 49 return NULL; 50 } 51 };
头一次一遍就AC……
个人思路：先计数，然后把起始点定好，都定在距离最后一个节点一样距离的地方。然后同步前进遍历即可。
改进：
为了节省计算，在计算链表长度的时候，顺便比较一下两个链表的尾节点是否一样，
若不一样，则不可能相交，直接可以返回NULL
leetcode - Intersection of Two Linked Lists
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原文地址：http://www.cnblogs.com/shnj/p/4563448.html
